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QuestionSelect tables that could represent a linear function based on the values of xx and f(x)f(x).

Studdy Solution

STEP 1

Assumptions1. A linear function can be represented in the form y=mx+cy = mx + c, where mm is the slope of the line and cc is the y-intercept. . For a table to represent a linear function, the difference in yy values divided by the difference in xx values (i.e., the slope) should be constant for all pairs of points.

STEP 2

Let's start with the first table. We calculate the slope between the first two points.
m1=k(10)−k(5)10−5m1 = \frac{k(10) - k(5)}{10 -5}

STEP 3

Substitute the values from the table into the equation.
m1=29−1410−5m1 = \frac{29 -14}{10 -5}

STEP 4

Calculate the slope between the first two points.
m1=15=3m1 = \frac{15}{} =3

STEP 5

Now, let's calculate the slope between the second and third points.
m2=k(20)−k(10)20−10m2 = \frac{k(20) - k(10)}{20 -10}

STEP 6

Substitute the values from the table into the equation.
m2=59−2920−10m2 = \frac{59 -29}{20 -10}

STEP 7

Calculate the slope between the second and third points.
m2=3010=3m2 = \frac{30}{10} =3

STEP 8

Since the slopes m1m1 and m2m2 are equal, the first table could represent a linear function.

STEP 9

Repeat the same process for the second table. Calculate the slope between the first two points.
m=h(5)−h()5−m = \frac{h(5) - h()}{5 -}

STEP 10

Substitute the values from the table into the equation.
m=35−105−0m = \frac{35 -10}{5 -0}

STEP 11

Calculate the slope between the first two points.
m=255=5m = \frac{25}{5} =5

STEP 12

Now, let's calculate the slope between the second and third points.
m2=h(10)−h(5)10−5m2 = \frac{h(10) - h(5)}{10 -5}

STEP 13

Substitute the values from the table into the equation.
m2=110−3510−5m2 = \frac{110 -35}{10 -5}

STEP 14

Calculate the slope between the second and third points.
m2=75=m2 = \frac{75}{} =

STEP 15

Since the slopes mm and m2m2 are not equal, the second table does not represent a linear function.

STEP 16

Repeat the same process for the third table. Calculate the slope between the first two points.
m=f(5)−f(0)5−0m = \frac{f(5) - f(0)}{5 -0}

STEP 17

Substitute the values from the table into the equation.
m=−(−2)5−0m = \frac{ - (-2)}{5 -0}

STEP 18

Calculate the slope between the first two points.
m=205=4m = \frac{20}{5} =4

STEP 19

Now, let's calculate the slope between the second and third points.
m=f(10)−f(5)10−5m = \frac{f(10) - f(5)}{10 -5}

STEP 20

Substitute the values from the table into the equation.
m=38−1810−5m = \frac{38 -18}{10 -5}

STEP 21

Calculate the slope between the second and third points.
m=205=4m = \frac{20}{5} =4

STEP 22

Since the slopes m1m1 and mm are equal, the third table could represent a linear function.
The tables that represent a linear function are the first and third tables.

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